Static equilibrium, diving board

In summary: So the forces are in the opposite direction, and hence your answers are negative. In summary, the force exerted on the support at x = 0 is -1017.5 N and the force acting on the other support is -1987.5 N.
  • #1
BikeSmoth
10
0

Homework Statement


A person who weighs 620 N stands at x = 5.00 m, right on the end of a long horizontal diving board that weighs 350 N. The diving board is held up by two supports, one at its left end at x = 0, and one at the point x = 2.00 m. (a) What is the force exerted on the support at x = 0? (b) What is the force acting on the other support? (Use positive to indicate an upward force, negative for a downward force.)


Homework Equations


sum T=0, T=FR
Mgp=wieght of the person
Mgb=weight of the board
FA=force at point a - the left most support
FB=Force at point b - the rightmost support

The Attempt at a Solution


I don't know what I am doing wrong, but apparently these answers are incorrect.
(a) sum T= 0 =-Mgp*Rp+(-Mgb)*Rb+FA*Ra
FA=[Mgp*Rp+Mgb*Rb]/Ra
all about point b
FA=[620N*3m+350N*.5m]/-2m
FA=-1017.5N
(b)FB=[Mgp*Rp+Mgb*Rb]/Rb
about point a
FB=[620*5+350*2.5]/2
FB=1987.5
 
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  • #2
BikeSmoth said:

Homework Statement


A person who weighs 620 N stands at x = 5.00 m, right on the end of a long horizontal diving board that weighs 350 N. The diving board is held up by two supports, one at its left end at x = 0, and one at the point x = 2.00 m. (a) What is the force exerted on the support at x = 0? (b) What is the force acting on the other support? (Use positive to indicate an upward force, negative for a downward force.)


Homework Equations


sum T=0, T=FR
Mgp=wieght of the person
Mgb=weight of the board
FA=force at point a - the left most support
FB=Force at point b - the rightmost support

The Attempt at a Solution


I don't know what I am doing wrong, but apparently these answers are incorrect.
(a) sum T= 0 =-Mgp*Rp+(-Mgb)*Rb+FA*Ra
FA=[Mgp*Rp+Mgb*Rb]/Ra
all about point b
FA=[620N*3m+350N*.5m]/-2m
FA=-1017.5N
(b)FB=[Mgp*Rp+Mgb*Rb]/Rb
about point a
FB=[620*5+350*2.5]/2
FB=1987.5
Your numbers are OK (might want to round them off), but read the question carefully. It asks for the force on the supports, not the force of the supports on the diving board (think of Newton 3).
 

Related to Static equilibrium, diving board

1. How does a diving board maintain static equilibrium?

A diving board maintains static equilibrium by having an equal amount of force acting on both sides of the board. This is achieved through the balance of the weight of the diver and the reaction force from the board's support.

2. What factors affect the static equilibrium of a diving board?

The factors that affect the static equilibrium of a diving board include the weight and position of the diver, the length and stiffness of the board, and the strength and stability of the support.

3. Can the static equilibrium of a diving board be altered?

Yes, the static equilibrium of a diving board can be altered by changing the position of the diver on the board, adjusting the length or stiffness of the board, or modifying the support structure.

4. Why is it important for a diving board to maintain static equilibrium?

Maintaining static equilibrium is important for a diving board to ensure the safety and stability of the diver. If the board is not in static equilibrium, it can lead to a loss of balance and potentially cause injury to the diver.

5. How is the center of mass related to the static equilibrium of a diving board?

The center of mass of the diving board and the center of mass of the diver must be aligned for the board to maintain static equilibrium. If the center of mass of the diver is not directly above the center of mass of the board, it can cause the board to tip and lose equilibrium.

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